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MU5PYN03 J. Bolmont
TABLE OF CONTENTS• Detection : some generalities• Interaction of photons
- How photons deposit their energy?- Application for gamma-ray spectroscopy‣ A very simple detector to understand a lot!
• Detection/spectroscopy of photons- Inorganic/Organic Scintillators- Photomultipliers and other photodetectors- Semiconductor detectors
• Applications- Detectors for gamma-ray astronomy: satellites and ground-based detectors
2
MU5PYN03 J. Bolmont
MY TOPICS OF INTEREST• Gamma-ray astronomy in the MeV-TeV range
3
Fermi (~20 MeV - 300 GeV)
H.E.S.S. (30 GeV - 10 TeV)
MU5PYN03 J. Bolmont
A FIRST WORD ON FERMI• Fermi-LAT : 20 MeV - 300 GeV
- Silicon Tracker- Calorimeter : CsI scintillators- Angular resolution ~0.15° for E > 10 GeV
• Fermi-GBM : 8 keV - 40 MeV- 12 NaI scintillators (a few keV - 1 MeV)- 2 BGO scintillators (150 keV - 30 MeV)- Angular resolution ~15°
4
SOME GENERALITIES
MU5PYN03 J. Bolmont
GENERALITIES• A detector allows to access different kinds of information:
- Number of particles (counters),- Energy of particles (spectrometer, calorimeter),- Energy deposit (dosimeter)
• There are different types of them- Gas detectors- Semiconductor detectors,- Scintillation detectors
• It is possible to combine several types of detectors to access different information: position, speed, etc.
6
MU5PYN03 J. Bolmont
DETECTION SYSTEM
• To be detected, a photon must interact in the detector‣ Concept of « event »
• The detector output consists in electric pulses for each interaction
• The electronics is used to convert these pulses and send it to…• … signal shaping and analysis, which allows to access the quantity
we look for
7
Source Detector Electronics Signal analysis
MU5PYN03 J. Bolmont
DETECTOR CHARACTERISTICS
• Each event produce a signal which contains some information about- The energy deposit during the interaction- The time of the interaction- The number of interactions per unit of time- The kind of interaction, etc.
• Sometimes it’s interesting to look at integrated values‣ E.g. by integrating the energies for a lot of interactions, we can get a
dose
8
MU5PYN03 J. Bolmont
DETECTOR CHARACTERISTICS• Linearity
- The measured quantity should vary linearly with the physical value we look for‣ E.g. SC+PM: the amplitude of the output electric pulse is
proportional to the energy deposit in the scintillator- Any non-linearity should be taken into account
• Energy resolution- The energy resolution is measured using spectral peaks. It depends
on the energy.- For a peak with centroid H0 and a width at half maximum FWHM:
9
(SC + PM)R =
FWHM
E=
p↵+ �E
ER ⌘ FWHM
H0<latexit sha1_base64="dSa8u67vpxILq1AaQQm8LKBtHaY=">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</latexit>
MU5PYN03 J. Bolmont
• Poor resolution (large value for R) leads to a broadening of spectral pulses and to a higher uncertainty for energy measurement
• Spectral peaks are Gaussians
10
DETECTOR CHARACTERISTICS
Ee
dN/dE
Eγ = hν Ee
dN/dE
Eγ = hν
FWHM = 2.35 �
H
H/2FWHM
MU5PYN03 J. Bolmont
• Same spectrum obtained with two detectors with different energy resolutions: NaI(Tl) (R ~ 7%) et Ge (R ~ 0.4%).
11
DETECTOR CHARACTERISTICS
MU5PYN03 J. Bolmont
EXERCICE 3.1• A gamma-ray spectrometer records peaks corresponding to two
different gamma-ray energies of 435 and 490 keV. What must be the energy resolution of the system (expressed as a percentage) in order just to distinguish these two peaks?
12
Go to the solution…
MU5PYN03 J. Bolmont
• Detection efficiency- In general, the geometry of a detector do not allow to collect
all incoming particles- We define‣ The absolute efficiency:
‣ The intrinsic efficiency:
13
DETECTOR CHARACTERISTICS
✏abs =nb. of pulses
nb. of particles emitted=
⌦
4⇡✏int
<latexit sha1_base64="vlNBOuHrMuByDf3NyQHmW1H0aIU=">AAADTnicjVFNT9wwEJ2EfmxpKUurnioVq6tKnFZZigQXJEQvvZVKXUAiCDne2cXCSSzbQUJRfgD/rreq/4CeeusN0bGblYClAkdJnt/Me/bMZFpJ65LkZxTPPXr85Gnn2fzzFwsvF7tLr3ZtWRmBQ1Gq0uxn3KKSBQ6ddAr3tUGeZwr3spNPPr53isbKsvjmzjQe5nxSyLEU3BF11D1PUVupCKY5d8cmr3lmG7bJ0rHhop6SRdZPWTlOma6URds0d0S4cVIotCnDXDqHo+aaz5ccJ7yp11ItGzZzpixcc9TtJf0kLDYLBi3oQbt2yu4PSGEEJQioIAeEAhxhBRwsPQcwgAQ0cYdQE2cIyRBHaGCetBVlIWVwYk/oO6HdQcsWtPeeNqgFnaLoNaRk8IE0JeUZwv40FuJVcPbs/7zr4Onvdkb/rPXKiXVwTOx9umnmQ3W+Fgdj2Ag1SKpJB8ZXJ1qXKnTF35xdq8qRgybO4xHFDWERlNM+s6CxoXbfWx7iFyHTs34v2twKfvlb0oAHt8c5C3ZX+4OP/dWva72t7XbUHXgL72GF5rkOW/AZdmBI3r+jN9G7aDn+Hv+JL+Orf6lx1Gpew4011/kLmeHF5g==</latexit>
✏int =nb. of pulses
nb. of particles hitting the detector<latexit sha1_base64="8B32PwHJmrbObAFYXJyf6UpAnS0=">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</latexit>
MU5PYN03 J. Bolmont
DETECTOR CHARACTERISTICS• Dead time
• The energy deposit of an incoming particle in a detector can be considered as instantaneous (O(ps - a few ns))
• The detector itself and the electronics can take a much longer time to deal with an event (O(tens of ns - hundreds of µs))‣ During that time, the detection system is « busy »‣ It’s not available to deal with a new event‣ « Dead time losses »
• The dead time should be as small as possible !
14
MU5PYN03 J. Bolmont
DETECTOR CHARACTERISTICS• Dead time
• Two extreme and idealized cases for counting applications:• Non-paralyzable (non-extendable)• Paralyzable (extendable)
15
Events in the detector
DeadLive
DeadLive
Time
Time
Time
Non-paralyzable
Paralyzableτ
✔ ✔ ✔ ✔
✔ ✔ ✔
« Real » detectors lie in between
MU5PYN03 J. Bolmont 16
DETECTOR CHARACTERISTICSO
bser
ved
rate
m
True rate n
m = n
Nonparalyzable
Paralyzable
1/τ
• Non-paralyzable: any event occurring during dead time τ is lost• Paralyzable: the system needs a rest of at least τ to be ready to
accept another event
MU5PYN03 J. Bolmont
DEADTIME - EXAMPLE
• Plot showing a distribution of the time difference between two consecutive events
• The sharp cut at 20 µs is due to the dead time (20 µs)
• Events for which the time difference is less than 20 µs are « ghosts »: false triggers due to electronics noise
17
H.E.S.S. - CT5
INTERACTION OF PHOTONS IN MATTER1 - CROSS-SECTION (QUICK RECAP)
MU5PYN03 J. Bolmont
• Given an element of volume S dx with n target atoms (« scattering centres ») per unit volume
‣ n in atoms/m3, ρ in g/m3, ℳ in g/mol (= A), NA in atoms/mol
• N particles enter the volume through a surface S
INTERACTION PROBABILITY
19
n =⇢
M NA
dxN(x) N(x+dx)
Surface S
σ
σσ
σσσ
σσ
MU5PYN03 J. Bolmont
• Each scattering center has a cross-sectional area σ• We consider a particle interacts if its path hits one of the areas σ. It is then
removed from the beam.• σ is the geometric reaction cross-section
- Expressed as a surface- Commonly used unit: the barn (1 b = 10-28 m2)- It depends on many factors (energy, particle type, target type, etc.) but it
does not depend on the experimental design
INTERACTION PROBABILITY
20
dxN(x) N(x+dx)
Surface S
σ
σσ
σσσ
σσ
MU5PYN03 J. Bolmont
INTERACTION PROBABILITY• For an homogeneous medium, diluted enough, the total surface
of interaction is the sum of cross-sections:
• The interaction probability is given by
• By definition, the interaction probability is the ratio between photons which interact and the total number of incoming photons entering the volume
21
s = nV �
P =s
S=
nV �
S=
nS dx�
S= n� dx
P = �dN
N= n� dx
MU5PYN03 J. Bolmont
NUMBER OF INTERACTIONS• It can be interesting to compute the number of interactions in a
given volume, knowing the incident flux of particles• We consider a flux of particles coming on particles at rest (target)
- We note ϕ the flux of incoming particles (m-2s-1)- We note n’ the number of target particles per unit area (m-2) =
n.dx- N’ is the number of particles which interacted (disappeared) per
unit of surface and time (m-2s-1) = ϕ.𝒫 with ‣ Then:
N 0 = � n0 �
P = n� dx
MU5PYN03 J. Bolmont
CROSS-SECTION - LUMINOSITY• The flux of particles (in a beam) can be expressed as a luminosity :
cm-2 s-1
• Integrating over time, we get integrated luminosity en cm-2 ou b-1
• If a detector integrateL = 1 fb-1 of p-p collisions over a year and if the cross-section for a given channel is σ = 1 mb, thenumber of events in that channel is σ/L = 1012
23
MU5PYN03 J. Bolmont
INTERACTION PROBABILITY• The resolution of the differential equation of slide 21 gives
• μ is the linear attenuation coefficient:• Its dimension is L-1
• The mean free path is λ = 1/μ• The mass attenuation coefficient is μ/ρ. It is usually given in cm2/g‣ Then it is possible to deduce that
• The half-value layer x1/2 is defined by• So
24
µ = n�
N(x1/2) = N0/2
N(x) = N0 e�n� x = N0 e
�µx
x1/2 =ln 2
µ
� =µ
⇢
A
NA(see slide 19)
MU5PYN03 J. Bolmont
USE FOR RADIOPROTECTION
• The thicker the screen is, the more attenuated the beam is• Hypotheses:
- « Narrow beam »- A photon which interacts is « lost »
• A more general expression: N(D) = N0 B(D,E) e-µD where B is the « buildup factor ».
25
N0 N(D) = N0 e-µD
Screen (µ)
D
Collimated beam
Small size detector
MU5PYN03 J. Bolmont
INTERACTION OF PHOTONS• Depending on its energy, a photon (X, γ) can interact through
different processes:- Photoelectric effect (absorption): γA → A+e-
- Compton scattering: γe- → γe-
- Pair production: γN → e+e-N- Rayleigh scattering, Thomson scattering, Hadron/Lepton pair
production…• Only photons which interact deposit their energy
26
Dominant for E > 10 eV
MU5PYN03 J. Bolmont
INTERACTION OF PHOTONS
• Absorption length, expressed in g cm-2 (≡ λρ)• Lead at 100 eV → λρ ~ 3.5x10-5 g cm-2 → λ ~ 30 nm
27
Photon energy
100
10
10 – 4
10 – 5
10 – 6
1
0.1
0.01
0.001
10 eV 100 eV 1 keV 10 keV 100 keV 1 MeV 10 MeV 100 MeV 1 GeV 10 GeV 100 GeV
Abs
orpt
ion
leng
th λ
(g/
cm2 )
Si
C
Fe Pb
H
Sn
MU5PYN03 J. Bolmont
INTERACTION OF PHOTONS• The total cross-section is
simply the sum of cross-sections for individual processesand sincewe have simply
28
�tot = �PE + �C + �CP
� =µ
⇢
A
NA
µtot = µPE + µC + µCPPhoton Energy
1 Mb
1 kb
1 b
10 mb10 eV 1 keV 1 MeV 1 GeV 100 GeV
(b) Lead (Z = 82)- experimental σtot
σp.e.
κe
Cro
ss se
ctio
n (b
arns
/ato
m)
Cro
ss se
ctio
n (b
arns
/ato
m)
10 mb
1 b
1 kb
1 Mb(a) Carbon (Z = 6)
σRayleigh
σg.d.r.
σCompton
σCompton
σRayleigh
κnuc
κnuc
κe
σp.e.
- experimental σtot
22 32. Passage of particles through matter
Photon Energy
1 Mb
1 kb
1 b
10 mb10 eV 1 keV 1 MeV 1 GeV 100 GeV
(b) Lead (Z = 82)- experimental σtot
σp.e.
κe
Cro
ss se
ctio
n (b
arns
/ato
m)
Cro
ss se
ctio
n (b
arns
/ato
m)
10 mb
1 b
1 kb
1 Mb(a) Carbon (Z = 6)
σRayleigh
σg.d.r.
σCompton
σCompton
σRayleigh
κnuc
κnuc
κe
σp.e.
- experimental σtot
Figure 32.15: Photon total cross sections as a function of energy in carbon and lead,showing the contributions of different processes [51]:
σp.e. = Atomic photoelectric effect (electron ejection, photon absorption)σRayleigh = Rayleigh (coherent) scattering–atom neither ionized nor excitedσCompton = Incoherent scattering (Compton scattering off an electron)
κnuc = Pair production, nuclear fieldκe = Pair production, electron field
σg.d.r. = Photonuclear interactions, most notably the Giant Dipole Resonance [52].In these interactions, the target nucleus is broken up.
Original figures through the courtesy of John H. Hubbell (NIST).
February 8, 2016 19:56
Photoelectric
Compton
Pair production
MU5PYN03 J. Bolmont
ENERGY DEPOSIT BY A X-RAY BEAM• A photon deposits its energy only if it interacts• The number of photons interacting per unit of length as a function of
depth isso, assuming µ does not depend on x (homogeneous material), and all photons have the same energy,
• Variation of the deposited energy as a function of distance traveled:
29
�dN(x)
dx
�dN(x)
dx= µN0 e
�µx ) dE(x)
dx= �E�
dN(x)
dx= µE� N0| {z }
E0
e�µx
dE
dx(x) = E0 e
�µx
MU5PYN03 J. Bolmont
ENERGY DEPOSIT BY A X-RAY BEAM• The energy deposit is maximum at the entrance in the material
and then decrease exponentially• By definition of x1/2, the energy deposited is divided by 2 at this
depth
30
x
At high energies, the way electrons interact will have an important
effect on overall energy deposit
dE/dxE0
E0/2
x1/2
Photons γ HEPhotons γ BE
Photons cannot be detected if they do not interact…
INTERACTION OF PHOTONS IN MATTER2 - HOW PHOTONS INTERACT (1)
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• If the photon has an energy hν greater
than ionization energy EI of an electron, it can be absorbed and the electron ejected with an energy Ee so that
• For energies above a few hundred keV, the electron carry most of the energy of the photon
• The vacancy left behind the electron is filled by another electron from another shell or by a free electron‣ Emission of X-rays or of Auger
electrons, usually quickly absorbed
32
γe-
E� = h ⌫ = Ee + EI
X
e- e-
e- Auger
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• Complexity !
33
γe-
X
e- e-
e- Auger
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• Most of the time, the electron is absorbed and all the energy of
the photon is deposited• Question: how the energy of electrons is distributed (for
monochromatic photons) ?
34
Ee
dN/dEγ
e-?
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• Most of the time, the electron is absorbed and all the energy of
the photon is deposited• Assuming a mono-energetic photon beam, all the electrons have
the same energy and the spectrum is a Dirac peak at Ee = hν.‣ Photo-peak or full-energy peak
35
Ee
dN/dE
hν
Photo-peak or full-energy peak γ
e-
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• In practice, if the interaction takes place on the edge of the
detector, the electron may escape without depositing all of its energy
36
Ee
dN/dE
hν
Photo-peak or full-energy peak
γ
e-
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• In practice, if the interaction takes place on the edge of the
detector, the electron may escape without depositing all of its energy
37
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Edep (MeV)
1
10
210
310
410
510
610
Edep in absorber - First Interaction = P.E. + e- exit (MeV)
Result from a simulation:662 keV photons interact only through p.e. effect(other interactions are
ignored)
« Perfect » resolution
98,7 %
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• For decreasing photon energies, the electrons ejected are from K
shell, L shell, etc.• The cross-section decrease fast with increasing photon energies• It increases quickly with target nuclei mass
38
At 100 keV :σPE(Fe) = 19 bσPE(Pb) = 1800 b
(From XCOM NIST database)
L3
L2
L1
K
�PE / Zn
E3.5�
<latexit sha1_base64="5KMWHQ1Te/nV2VYaWHOyug8gvIw=">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</latexit>
n varies with energy between 4 and 5
MU5PYN03 J. Bolmont
PHOTOELECTRIC ABSORPTION• The breaks correspond to ionization energies for K electrons, L
electrons, etc.
39
Photon energy is too low to eject K electrons. K
electrons are not available.
Photon energy is high enough to eject
K electrons.K electrons are
available.
(From XCOM NIST database)
MU5PYN03 J. Bolmont
COMPTON SCATTERING• Scattering of a photon on a lightly bounded
electron• The photon is scattered and loose energy,
which is given to the recoil electron• We can show that
• The energy of the electron is• Compton scattering dominates for Eν
between ~10-100 keV and ~10-100 MeV
40
γ
γ’
e-
θ
E�0 =E�
1 + E�
me c2 (1� cos ✓)
Ee = E� � E�0
The scattered photon « lives its life » !It can interact again or not !
MU5PYN03 J. Bolmont
COMPTON SCATTERING
• The energy of the scattered photon, and so the one of the electron, varies continuously when θ varies between 0 and π- For θ = 0, Eγ’ = Eγ, and Ee = 0- For θ = π, and in this case,
• The difference between the maximum energy of the recoil electron and the energy of the incoming photon is
41
EC ⌘ E� � Ee(✓ = ⇡) =E�
1 + 2 E�
me c2
Ee(✓ = ⇡) = E�
2 E�
me c2
1 + 2 E�
me c2
E�0(✓ = ⇡) =E�
1 + 2 E�
me c2
E�0 =E�
1 + E�
me c2 (1� cos ✓)
MU5PYN03 J. Bolmont
EXERCICE 3.1• We want the peaks to be separated by at least 1 FWHM• The separation is 55 keV so we need a resolution of 55/435 =
12.6% or less at 435 keV and 11.2% or less at 490 keV.• So, we need a resolution of 11.2% or less.
42
Back to the exercice...
MU5PYN03 J. Bolmont
COMPTON SCATTERING• Scattering of a photon on a lightly bounded
electron• The photon is scattered and loose energy,
which is given to the recoil electron• We can show that
• The energy of the electron is• Compton scattering dominates for Eν
between ~10-100 keV and ~10-100 MeV
43
γ
γ’
e-
θ
E�0 =E�
1 + E�
me c2 (1� cos ✓)
Ee = E� � E�0
The scattered photon « lives its life » !It can interact again or not !
MU5PYN03 J. Bolmont
COMPTON SCATTERING• Electron spectrum is continuous• Since really often Eγ >> mec2/2, EC ≈ mec2/2 ≈ 0.256 MeV
44
Ee
dN/dEθ = 0 θ = π
hν
Compton continuum
EC
Compton edge
γ
e-
EC ⌘ E� � Ee(✓ = ⇡) =E�
1 + 2 E�
me c2
MU5PYN03 J. Bolmont
COMPTON SCATTERING• Shape of Compton continuum for various gamma-ray energies• Variation of the energy of the scattered photon as a function of the angle
45
'Y = -y Ray energy
1.5 'Y = 0.4 moc2 _ Electron energy t -\ m c2
\ 0 \ \ '0.5
\ \
t \ \
\ \
N 1.0 E \ (,,)
\ N 1 0 .-X
cl 1.1 ""= ""= ?'-- 2
' 3 ,45
.1°20 50
0 0 0.2 0.4 0.6 0.8 1.0
€/-y >
Figure 10.1 Shape of the Compt?n. co~tinuum for various gamma-ray energies. (From s. M. Shafroth ( ed. ), Sczntzllation Spectroscopy of Gamma Radiation. Copyright 1964 by Gordon & Breach, Inc. By permission of the publisher.)
-> Q)
-> O') ... Cl) C: Cl)
C 0 ... 0 .c o..
2.80
2.40
1.60
1.20
0.80
0.40
Compton scattered photon energy as a function of scatter ing angle
o~-+---+---+---+--+---+---t----ti--~~-..,__,.___. 0 30 60 90 120 150 180
Scattering angle (degrees)
Figure 10. 7 Variation of scattered gamma .. ray energy with scattering angle.
~ Cste ~ 200 keVWhatever the initial
energy
MU5PYN03 J. Bolmont
COMPTON SCATTERING• Electron spectrum is continuous• Since really often Eγ >> mec2/2, EC ≈ mec2/2 ≈ 0.256 MeV
46
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Edep (MeV)
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
Edep in absorber - First Interaction = Compton (MeV)
hν
Result from a simulation:662 keV photons interact only through Compton
scattering(other interactions are
ignored)
« Perfect » resolution + only one scattering is
allowed
MU5PYN03 J. Bolmont
e+e- PAIR PRODUCTION
• The photon disappear creating a electron-positron pair• This process can only occur in the EM field of a nuclei (or of an e-)• Threshold energy: • The energy in excess of ~2mec2 is transferred as kinetic energy of
the e+ and e-:
47
γ
e-
e+
Ee+ + Ee� = E� � 2mec2
E� � 2me c2
✓1 +
me
mN
◆
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MU5PYN03 J. Bolmont
ANNIHILATION
• Positons travel through matter losing their energy‣ Remember part 2 of this course!
• After a short time (~few ps), when they are (almost) at rest, they interact with electrons
• e+ form a bound state with e-, called positronium• Para-positronium (2, 4, 6… photons produced, T~125 ps) or• Ortho-positronium (3, 5, 7… photons produced, T~142 ns)
48
e+ + e� ! 2 �
e-
e+ γγ
MU5PYN03 J. Bolmont
ANNIHILATION• Energy conservation :• Momentum conservation
‣ It’s impossible to produce only one particle !‣ In case of two photons, they are emitted at 180° from each other
• Energy and momentum of the photon:
• So,• Energy conservation gives:
49
E� = h ⌫ p� = E�/c
�!P�1 +
�!P�2 =
�!0 ) E�1 = E�2
E�1 = E�2 = me c2 = 511 keV
�!P�1 +
�!P�2 =
�!0
me c2 +me c
2 = E�1 + E�2
MU5PYN03 J. Bolmont
RELATIVE IMPORTANCE OF THE THREE MODES OF INTERACTION
• The relative importance of the three modes depends on the Z of target material and on the energy of incoming gamma-rays
• On the plot below, the lines are for equal cross sections
50
MU5PYN03 J. Bolmont
CONTRIBUTIONS TO THE ABSORPTION COEFFICIENT
• Each of the effects seen above contribute to the total cross section
• For a composite material,
51
NaI
✓µ
⇢
◆
tot
=X
i
wi
✓µ
⇢
◆
i
X
i
wi = 1
http://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html
MU5PYN03 J. Bolmont
CONTRIBUTIONS TO THE ABSORPTION COEFFICIENT
52
http://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html
MU5PYN03 J. Bolmont
EXERCICE 4.1• Let’s imagine a photon with hν >> 2 mec2 interact through pair
production, that the electron is absorbed, and the positon annihilates.
• Draw the possible cases.• What deposited energy(ies) can be measured ?
53
Ee
dN/dEhν
MU5PYN03 J. Bolmont
e+e- PAIR PRODUCTION• The probability of pair production increase above • Peaks in the spectrum are difficult to observe when • The positon quickly annihilates to give (mostly) two 511 keV photons
which may interact, or not.
54
Ee
dN/dEhν
hν - 2mec2 hν - mec2
E� < 4mec2
E� > 2mec2
γ
e-e+
The two photons escape: « double escape ».If only one of them escapes: « single escape ».
« Double escape » peak
« Single escape » peak
(Solution for Exercice 4.2…)
MU5PYN03 J. Bolmont
TOTAL ABSORPTION• Whatever the type of the first interaction, or the exact way the
event develops, the deposited energy is the incident energy !‣ « Full energy peak »
55
γe- γ
e-
e-
e-
γ
e- e-
e-e-
e+
« Total absorption » - « Full energy peak »
MU5PYN03 J. Bolmont
SMALL DETECTOR• Scattered photons, or photons
produced by e+ annihilation can escape the detector
• If they escape (before interacting), their energy is not deposited‣ We don’t measure all the
incident energy !‣ Missing energy !
56
γe-
γ
e-γ
e-e+
MU5PYN03 J. Bolmont
SPECTRUM FOR A SMALL DETECTOR
57
Ee
dN/dE
Eγ = hνhν - 2mec2
Ee
dN/dE
Eγ = hν
Eγ < 2mec2
Eγ >> 2mec2
hν - mec2
WARNING !
Perfect resolution !No multiple scattering !Amplitudes of the peaks are not to scale !
MU5PYN03 J. Bolmont
SPECTRUM FOR A SMALL DETECTOR
58
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Edep (MeV)
10
210
310
410
510
610
Edep in absorber (MeV)
Result from a simulation:662 keV photons interact in
a small size detector. All three interaction modes are
included.
« Perfect » resolution
